Skip to main content
Super TutorResources
Chapter 12 of 14
QuestionsNotesFlashcardsSummaryFormulasQuizConceptsSyllabusPlan
NCERT Solutions

Limits and Derivatives

Bihar Board · Class 11 · Mathematics

NCERT Solutions for Limits and Derivatives — Bihar Board Class 11 Mathematics.

84 questions22 flashcards5 concepts

Interactive on Super Tutor

Studying Limits and Derivatives? Get the full interactive chapter.

Quizzes, flashcards, AI doubt-solver and a step-by-step study plan — built for ncert solutions and more.

Try Super Tutor — Free

1,000+ Class 11 students started this chapter today

73 Questions Solved · 3 Sections

Exercise 12.1

1lim⁡x→3(x+3)\lim_{x\to 3}(x + 3)Show solution
Given: lim⁡x→3(x+3)\lim_{x\to 3}(x + 3)

Concept: For a polynomial function, the limit is found by direct substitution.

Working:
lim⁡x→3(x+3)=3+3=6\lim_{x\to 3}(x + 3) = 3 + 3 = 6

Answer: 66
2lim⁡x→π(x−227)\lim_{x\to \pi}\left(x - \dfrac{22}{7}\right)Show solution
Given: lim⁡x→π(x−227)\lim_{x\to \pi}\left(x - \dfrac{22}{7}\right)

Concept: Direct substitution for a polynomial/linear function.

Working:
lim⁡x→π(x−227)=π−227\lim_{x\to \pi}\left(x - \frac{22}{7}\right) = \pi - \frac{22}{7}

Answer: π−227\pi - \dfrac{22}{7}
3lim⁡r→1πr2\lim_{r\to 1}\pi r^2Show solution
Given: lim⁡r→1πr2\lim_{r\to 1}\pi r^2

Concept: Direct substitution.

Working:
lim⁡r→1πr2=π(1)2=π\lim_{r\to 1}\pi r^2 = \pi(1)^2 = \pi

Answer: π\pi
4lim⁡x→44x+3x−2\lim_{x\to 4}\dfrac{4x + 3}{x - 2}Show solution
Given: lim⁡x→44x+3x−2\lim_{x\to 4}\dfrac{4x + 3}{x - 2}

Concept: Direct substitution (denominator ≠0\neq 0 at x=4x = 4).

Working:
lim⁡x→44x+3x−2=4(4)+34−2=16+32=192\lim_{x\to 4}\frac{4x + 3}{x - 2} = \frac{4(4) + 3}{4 - 2} = \frac{16 + 3}{2} = \frac{19}{2}

Answer: 192\dfrac{19}{2}
5lim⁡x→−1x10+x5+1x−1\lim_{x\to -1}\dfrac{x^{10} + x^5 + 1}{x - 1}Show solution
Given: lim⁡x→−1x10+x5+1x−1\lim_{x\to -1}\dfrac{x^{10} + x^5 + 1}{x - 1}

Concept: Direct substitution (denominator ≠0\neq 0 at x=−1x = -1).

Working:
lim⁡x→−1x10+x5+1x−1=(−1)10+(−1)5+1−1−1=1−1+1−2=1−2=−12\lim_{x\to -1}\frac{x^{10} + x^5 + 1}{x - 1} = \frac{(-1)^{10} + (-1)^5 + 1}{-1 - 1} = \frac{1 - 1 + 1}{-2} = \frac{1}{-2} = -\frac{1}{2}

Answer: −12-\dfrac{1}{2}
6lim⁡x→0(x+1)5−1x\lim_{x\to 0}\dfrac{(x + 1)^5 - 1}{x}Show solution
Given: lim⁡x→0(x+1)5−1x\lim_{x\to 0}\dfrac{(x + 1)^5 - 1}{x}

Concept: Use the standard limit lim⁡x→axn−anx−a=nan−1\lim_{x\to a}\dfrac{x^n - a^n}{x - a} = na^{n-1}.

Let y=x+1y = x + 1, so as x→0x \to 0, y→1y \to 1.

Working:
lim⁡x→0(x+1)5−1x=lim⁡y→1y5−15y−1=5⋅15−1=5×1=5\lim_{x\to 0}\frac{(x+1)^5 - 1}{x} = \lim_{y\to 1}\frac{y^5 - 1^5}{y - 1} = 5 \cdot 1^{5-1} = 5 \times 1 = 5

Answer: 55
7lim⁡x→23x2−x−10x2−4\lim_{x\to 2}\dfrac{3x^2 - x - 10}{x^2 - 4}Show solution
Given: lim⁡x→23x2−x−10x2−4\lim_{x\to 2}\dfrac{3x^2 - x - 10}{x^2 - 4}

At x=2x = 2: numerator =12−2−10=0= 12 - 2 - 10 = 0, denominator =0= 0. So we factorise.

Factorising the numerator:
3x2−x−10=(x−2)(3x+5)3x^2 - x - 10 = (x - 2)(3x + 5)

Factorising the denominator:
x2−4=(x−2)(x+2)x^2 - 4 = (x - 2)(x + 2)

Working:
lim⁡x→2(x−2)(3x+5)(x−2)(x+2)=lim⁡x→23x+5x+2=3(2)+52+2=114\lim_{x\to 2}\frac{(x-2)(3x+5)}{(x-2)(x+2)} = \lim_{x\to 2}\frac{3x+5}{x+2} = \frac{3(2)+5}{2+2} = \frac{11}{4}

Answer: 114\dfrac{11}{4}
8lim⁡x→3x4−812x2−5x−3\lim_{x\to 3}\dfrac{x^4 - 81}{2x^2 - 5x - 3}Show solution
Given: lim⁡x→3x4−812x2−5x−3\lim_{x\to 3}\dfrac{x^4 - 81}{2x^2 - 5x - 3}

At x=3x = 3: numerator =81−81=0= 81 - 81 = 0, denominator =18−15−3=0= 18 - 15 - 3 = 0. Factorise.

Factorising numerator:
x4−81=(x2−9)(x2+9)=(x−3)(x+3)(x2+9)x^4 - 81 = (x^2 - 9)(x^2 + 9) = (x-3)(x+3)(x^2+9)

Factorising denominator:
2x2−5x−3=(x−3)(2x+1)2x^2 - 5x - 3 = (x - 3)(2x + 1)

Working:
lim⁡x→3(x−3)(x+3)(x2+9)(x−3)(2x+1)=lim⁡x→3(x+3)(x2+9)2x+1=(6)(18)7=1087\lim_{x\to 3}\frac{(x-3)(x+3)(x^2+9)}{(x-3)(2x+1)} = \lim_{x\to 3}\frac{(x+3)(x^2+9)}{2x+1} = \frac{(6)(18)}{7} = \frac{108}{7}

Answer: 1087\dfrac{108}{7}
9lim⁡x→0ax+bcx+1\lim_{x\to 0}\dfrac{ax + b}{cx + 1}Show solution
Given: lim⁡x→0ax+bcx+1\lim_{x\to 0}\dfrac{ax + b}{cx + 1}

Concept: Direct substitution (denominator =1≠0= 1 \neq 0 at x=0x = 0).

Working:
lim⁡x→0ax+bcx+1=a(0)+bc(0)+1=b1=b\lim_{x\to 0}\frac{ax + b}{cx + 1} = \frac{a(0) + b}{c(0) + 1} = \frac{b}{1} = b

Answer: bb
10lim⁡z→1z3−1z6−1\lim_{z\to 1}\dfrac{z^3 - 1}{z^6 - 1}Show solution
Given: lim⁡z→1z3−1z6−1\lim_{z\to 1}\dfrac{z^3 - 1}{z^6 - 1}

At z=1z = 1: both numerator and denominator are 00. Factorise.

Working:
lim⁡z→1z3−1z6−1=lim⁡z→1z3−1(z3)2−12=lim⁡z→1z3−1(z3−1)(z3+1)\lim_{z\to 1}\frac{z^3 - 1}{z^6 - 1} = \lim_{z\to 1}\frac{z^3 - 1}{(z^3)^2 - 1^2} = \lim_{z\to 1}\frac{z^3 - 1}{(z^3 - 1)(z^3 + 1)}
=lim⁡z→11z3+1=11+1=12= \lim_{z\to 1}\frac{1}{z^3 + 1} = \frac{1}{1 + 1} = \frac{1}{2}

Answer: 12\dfrac{1}{2}
11lim⁡x→1ax2+bx+ccx2+bx+a,a+b+c≠0\lim_{x\to 1}\dfrac{ax^2 + bx + c}{cx^2 + bx + a},\quad a + b + c \neq 0Show solution
Given: lim⁡x→1ax2+bx+ccx2+bx+a\lim_{x\to 1}\dfrac{ax^2 + bx + c}{cx^2 + bx + a}, where a+b+c≠0a + b + c \neq 0.

Concept: At x=1x = 1, numerator =a+b+c≠0= a + b + c \neq 0 and denominator =c+b+a=a+b+c≠0= c + b + a = a + b + c \neq 0. So direct substitution applies.

Working:
lim⁡x→1ax2+bx+ccx2+bx+a=a(1)+b(1)+cc(1)+b(1)+a=a+b+ca+b+c=1\lim_{x\to 1}\frac{ax^2 + bx + c}{cx^2 + bx + a} = \frac{a(1) + b(1) + c}{c(1) + b(1) + a} = \frac{a + b + c}{a + b + c} = 1

Answer: 11
12lim⁡x→−21x+12x+2\lim_{x\to -2}\dfrac{\dfrac{1}{x} + \dfrac{1}{2}}{x + 2}Show solution
Given: lim⁡x→−21x+12x+2\lim_{x\to -2}\dfrac{\dfrac{1}{x} + \dfrac{1}{2}}{x + 2}

At x=−2x = -2: numerator =−12+12=0= -\frac{1}{2} + \frac{1}{2} = 0, denominator =0= 0. Simplify.

Working:
1x+12x+2=2+x2xx+2=x+22x(x+2)=12x\frac{\frac{1}{x} + \frac{1}{2}}{x + 2} = \frac{\frac{2 + x}{2x}}{x + 2} = \frac{x + 2}{2x(x + 2)} = \frac{1}{2x}

lim⁡x→−212x=12(−2)=−14\lim_{x\to -2}\frac{1}{2x} = \frac{1}{2(-2)} = -\frac{1}{4}

Answer: −14-\dfrac{1}{4}
13lim⁡x→0sin⁡axbx\lim_{x\to 0}\dfrac{\sin ax}{bx}Show solution
Given: lim⁡x→0sin⁡axbx\lim_{x\to 0}\dfrac{\sin ax}{bx}

Concept: Use the standard limit lim⁡θ→0sin⁡θθ=1\lim_{\theta\to 0}\dfrac{\sin\theta}{\theta} = 1.

Working:
lim⁡x→0sin⁡axbx=lim⁡x→0sin⁡axax⋅axbx=lim⁡x→0sin⁡axax⋅ab\lim_{x\to 0}\frac{\sin ax}{bx} = \lim_{x\to 0}\frac{\sin ax}{ax} \cdot \frac{ax}{bx} = \lim_{x\to 0}\frac{\sin ax}{ax} \cdot \frac{a}{b}
=1⋅ab=ab= 1 \cdot \frac{a}{b} = \frac{a}{b}

Answer: ab\dfrac{a}{b}
14lim⁡x→0sin⁡axsin⁡bx,a,b≠0\lim_{x\to 0}\dfrac{\sin ax}{\sin bx},\quad a, b \neq 0Show solution
Given: lim⁡x→0sin⁡axsin⁡bx\lim_{x\to 0}\dfrac{\sin ax}{\sin bx}, a,b≠0a, b \neq 0.

Concept: Use lim⁡θ→0sin⁡θθ=1\lim_{\theta\to 0}\dfrac{\sin\theta}{\theta} = 1.

Working:
lim⁡x→0sin⁡axsin⁡bx=lim⁡x→0sin⁡axax⋅axsin⁡bxbx⋅bx=lim⁡x→0sin⁡axaxlim⁡x→0sin⁡bxbx⋅ab=11⋅ab=ab\lim_{x\to 0}\frac{\sin ax}{\sin bx} = \lim_{x\to 0}\frac{\dfrac{\sin ax}{ax} \cdot ax}{\dfrac{\sin bx}{bx} \cdot bx} = \frac{\lim_{x\to 0}\dfrac{\sin ax}{ax}}{\lim_{x\to 0}\dfrac{\sin bx}{bx}} \cdot \frac{a}{b} = \frac{1}{1} \cdot \frac{a}{b} = \frac{a}{b}

Answer: ab\dfrac{a}{b}
15lim⁡x→πsin⁡(π−x)π(π−x)\lim_{x\to \pi}\dfrac{\sin(\pi - x)}{\pi(\pi - x)}Show solution
Given: lim⁡x→πsin⁡(π−x)π(π−x)\lim_{x\to \pi}\dfrac{\sin(\pi - x)}{\pi(\pi - x)}

Concept: Let y=π−xy = \pi - x. As x→πx \to \pi, y→0y \to 0.

Working:
lim⁡x→πsin⁡(π−x)π(π−x)=lim⁡y→0sin⁡yπy=1πlim⁡y→0sin⁡yy=1π⋅1=1π\lim_{x\to \pi}\frac{\sin(\pi - x)}{\pi(\pi - x)} = \lim_{y\to 0}\frac{\sin y}{\pi y} = \frac{1}{\pi}\lim_{y\to 0}\frac{\sin y}{y} = \frac{1}{\pi} \cdot 1 = \frac{1}{\pi}

Answer: 1π\dfrac{1}{\pi}
16lim⁡x→0cos⁡xπ−x\lim_{x\to 0}\dfrac{\cos x}{\pi - x}Show solution
Given: lim⁡x→0cos⁡xπ−x\lim_{x\to 0}\dfrac{\cos x}{\pi - x}

Concept: Direct substitution (denominator =π≠0= \pi \neq 0 at x=0x = 0).

Working:
lim⁡x→0cos⁡xπ−x=cos⁡0π−0=1π\lim_{x\to 0}\frac{\cos x}{\pi - x} = \frac{\cos 0}{\pi - 0} = \frac{1}{\pi}

Answer: 1π\dfrac{1}{\pi}
17lim⁡x→0cos⁡2x−1cos⁡x−1\lim_{x\to 0}\dfrac{\cos 2x - 1}{\cos x - 1}Show solution
Given: lim⁡x→0cos⁡2x−1cos⁡x−1\lim_{x\to 0}\dfrac{\cos 2x - 1}{\cos x - 1}

Concept: Use 1−cos⁡θ=2sin⁡2θ21 - \cos\theta = 2\sin^2\dfrac{\theta}{2}.

Working:
cos⁡2x−1=−2sin⁡2x,cos⁡x−1=−2sin⁡2x2\cos 2x - 1 = -2\sin^2 x, \quad \cos x - 1 = -2\sin^2\frac{x}{2}

lim⁡x→0−2sin⁡2x−2sin⁡2x2=lim⁡x→0sin⁡2xsin⁡2x2\lim_{x\to 0}\frac{-2\sin^2 x}{-2\sin^2\frac{x}{2}} = \lim_{x\to 0}\frac{\sin^2 x}{\sin^2\frac{x}{2}}

Using sin⁡x=2sin⁡x2cos⁡x2\sin x = 2\sin\dfrac{x}{2}\cos\dfrac{x}{2}:
=lim⁡x→04sin⁡2x2cos⁡2x2sin⁡2x2=lim⁡x→04cos⁡2x2=4cos⁡20=4×1=4= \lim_{x\to 0}\frac{4\sin^2\frac{x}{2}\cos^2\frac{x}{2}}{\sin^2\frac{x}{2}} = \lim_{x\to 0}4\cos^2\frac{x}{2} = 4\cos^2 0 = 4 \times 1 = 4

Answer: 44
18lim⁡x→0ax+xcos⁡xbsin⁡x\lim_{x\to 0}\dfrac{ax + x\cos x}{b\sin x}Show solution
Given: lim⁡x→0ax+xcos⁡xbsin⁡x\lim_{x\to 0}\dfrac{ax + x\cos x}{b\sin x}

Working:
lim⁡x→0x(a+cos⁡x)bsin⁡x=lim⁡x→0xsin⁡x⋅a+cos⁡xb\lim_{x\to 0}\frac{x(a + \cos x)}{b\sin x} = \lim_{x\to 0}\frac{x}{\sin x} \cdot \frac{a + \cos x}{b}
=lim⁡x→01sin⁡xx⋅a+cos⁡xb=11⋅a+cos⁡0b=a+1b= \lim_{x\to 0}\frac{1}{\dfrac{\sin x}{x}} \cdot \frac{a + \cos x}{b} = \frac{1}{1} \cdot \frac{a + \cos 0}{b} = \frac{a + 1}{b}

Answer: a+1b\dfrac{a + 1}{b}
19lim⁡x→0xsec⁡x\lim_{x\to 0}x\sec xShow solution
Given: lim⁡x→0xsec⁡x\lim_{x\to 0}x\sec x

Concept: Direct substitution.

Working:
lim⁡x→0xsec⁡x=lim⁡x→0xcos⁡x=0cos⁡0=01=0\lim_{x\to 0}x\sec x = \lim_{x\to 0}\frac{x}{\cos x} = \frac{0}{\cos 0} = \frac{0}{1} = 0

Answer: 00
20lim⁡x→0sin⁡ax+bxax+sin⁡bx,a,b,a+b≠0\lim_{x\to 0}\dfrac{\sin ax + bx}{ax + \sin bx},\quad a, b, a+b \neq 0Show solution
Given: lim⁡x→0sin⁡ax+bxax+sin⁡bx\lim_{x\to 0}\dfrac{\sin ax + bx}{ax + \sin bx}

Concept: Divide numerator and denominator by xx.

Working:
lim⁡x→0sin⁡axx+ba+sin⁡bxx=lim⁡x→0a⋅sin⁡axax+ba+b⋅sin⁡bxbx\lim_{x\to 0}\frac{\dfrac{\sin ax}{x} + b}{a + \dfrac{\sin bx}{x}} = \lim_{x\to 0}\frac{a\cdot\dfrac{\sin ax}{ax} + b}{a + b\cdot\dfrac{\sin bx}{bx}}
=a⋅1+ba+b⋅1=a+ba+b=1= \frac{a \cdot 1 + b}{a + b \cdot 1} = \frac{a + b}{a + b} = 1

Answer: 11
21lim⁡x→0(csc⁡x−cot⁡x)\lim_{x\to 0}(\csc x - \cot x)Show solution
Given: lim⁡x→0(csc⁡x−cot⁡x)\lim_{x\to 0}(\csc x - \cot x)

Working:
lim⁡x→0(1sin⁡x−cos⁡xsin⁡x)=lim⁡x→01−cos⁡xsin⁡x\lim_{x\to 0}\left(\frac{1}{\sin x} - \frac{\cos x}{\sin x}\right) = \lim_{x\to 0}\frac{1 - \cos x}{\sin x}

Multiply numerator and denominator by xx:
=lim⁡x→01−cos⁡xx⋅xsin⁡x=0⋅1=0= \lim_{x\to 0}\frac{1 - \cos x}{x} \cdot \frac{x}{\sin x} = 0 \cdot 1 = 0

(Using standard limits: lim⁡x→01−cos⁡xx=0\lim_{x\to 0}\dfrac{1-\cos x}{x} = 0 and lim⁡x→0xsin⁡x=1\lim_{x\to 0}\dfrac{x}{\sin x} = 1.)

Answer: 00
22lim⁡x→π2tan⁡2xx−π2\lim_{x\to \frac{\pi}{2}}\dfrac{\tan 2x}{x - \dfrac{\pi}{2}}Show solution
Given: lim⁡x→π2tan⁡2xx−π2\lim_{x\to \frac{\pi}{2}}\dfrac{\tan 2x}{x - \dfrac{\pi}{2}}

Concept: Let y=x−π2y = x - \dfrac{\pi}{2}, so as x→π2x \to \dfrac{\pi}{2}, y→0y \to 0, and x=y+π2x = y + \dfrac{\pi}{2}.

Working:
tan⁡2x=tan⁡2(y+π2)=tan⁡(2y+π)=tan⁡2y\tan 2x = \tan 2\left(y + \frac{\pi}{2}\right) = \tan(2y + \pi) = \tan 2y

lim⁡y→0tan⁡2yy=lim⁡y→0sin⁡2yycos⁡2y=lim⁡y→0sin⁡2y2y⋅2cos⁡2y=1⋅21=2\lim_{y\to 0}\frac{\tan 2y}{y} = \lim_{y\to 0}\frac{\sin 2y}{y \cos 2y} = \lim_{y\to 0}\frac{\sin 2y}{2y} \cdot \frac{2}{\cos 2y} = 1 \cdot \frac{2}{1} = 2

Answer: 22
23Find lim⁡x→0f(x)\lim_{x\to 0}f(x) and lim⁡x→1f(x)\lim_{x\to 1}f(x), where f(x)={2x+3,amp;x≤03(x+1),amp;xgt;0f(x) = \begin{cases} 2x + 3, & x\leq 0 \\ 3(x + 1), & x > 0 \end{cases}Show solution
Given: f(x)={2x+3,amp;x≤03(x+1),amp;xgt;0f(x) = \begin{cases} 2x + 3, & x\leq 0 \\ 3(x + 1), & x > 0 \end{cases}

Finding lim⁡x→0f(x)\lim_{x\to 0}f(x):

Left-hand limit (LHL):
lim⁡x→0−f(x)=lim⁡x→0−(2x+3)=2(0)+3=3\lim_{x\to 0^-}f(x) = \lim_{x\to 0^-}(2x + 3) = 2(0) + 3 = 3

Right-hand limit (RHL):
lim⁡x→0+f(x)=lim⁡x→0+3(x+1)=3(0+1)=3\lim_{x\to 0^+}f(x) = \lim_{x\to 0^+}3(x + 1) = 3(0 + 1) = 3

Since LHL == RHL =3= 3:
lim⁡x→0f(x)=3\lim_{x\to 0}f(x) = 3

Finding lim⁡x→1f(x)\lim_{x\to 1}f(x):

For xx near 11 (both sides), x > 0, so f(x)=3(x+1)f(x) = 3(x+1).

LHL:
lim⁡x→1−f(x)=3(1+1)=6\lim_{x\to 1^-}f(x) = 3(1 + 1) = 6

RHL:
lim⁡x→1+f(x)=3(1+1)=6\lim_{x\to 1^+}f(x) = 3(1 + 1) = 6

Since LHL == RHL =6= 6:
lim⁡x→1f(x)=6\lim_{x\to 1}f(x) = 6
24Find lim⁡x→1f(x)\lim_{x\to 1}f(x), where f(x)={x2−1,amp;x≤1−x2−1,amp;xgt;1f(x) = \begin{cases} x^2 - 1, & x \leq 1 \\ -x^2 - 1, & x > 1 \end{cases}Show solution
Given: f(x)={x2−1,amp;x≤1−x2−1,amp;xgt;1f(x) = \begin{cases} x^2 - 1, & x \leq 1 \\ -x^2 - 1, & x > 1 \end{cases}

LHL:
lim⁡x→1−f(x)=lim⁡x→1−(x2−1)=1−1=0\lim_{x\to 1^-}f(x) = \lim_{x\to 1^-}(x^2 - 1) = 1 - 1 = 0

RHL:
lim⁡x→1+f(x)=lim⁡x→1+(−x2−1)=−1−1=−2\lim_{x\to 1^+}f(x) = \lim_{x\to 1^+}(-x^2 - 1) = -1 - 1 = -2

Since LHL ≠\neq RHL, lim⁡x→1f(x)\lim_{x\to 1}f(x) does not exist.
25Evaluate lim⁡x→0f(x)\lim_{x\to 0}f(x), where f(x)={∣x∣x,amp;x≠00,amp;x=0f(x) = \begin{cases} \dfrac{|x|}{x}, & x\neq 0\\ 0, & x = 0 \end{cases}Show solution
Given: f(x)={∣x∣x,amp;x≠00,amp;x=0f(x) = \begin{cases} \dfrac{|x|}{x}, & x\neq 0\\ 0, & x = 0 \end{cases}

For x > 0: ∣x∣=x|x| = x, so f(x)=xx=1f(x) = \dfrac{x}{x} = 1.

For x < 0: ∣x∣=−x|x| = -x, so f(x)=−xx=−1f(x) = \dfrac{-x}{x} = -1.

LHL:
lim⁡x→0−f(x)=−1\lim_{x\to 0^-}f(x) = -1

RHL:
lim⁡x→0+f(x)=1\lim_{x\to 0^+}f(x) = 1

Since LHL ≠\neq RHL, lim⁡x→0f(x)\lim_{x\to 0}f(x) does not exist.
26Find lim⁡x→0f(x)\lim_{x\to 0}f(x), where f(x)={x∣x∣,amp;x≠00,amp;x=0f(x) = \begin{cases} \dfrac{x}{|x|}, & x\neq 0\\ 0, & x = 0 \end{cases}Show solution
Given: f(x)={x∣x∣,amp;x≠00,amp;x=0f(x) = \begin{cases} \dfrac{x}{|x|}, & x\neq 0\\ 0, & x = 0 \end{cases}

For x > 0: f(x)=xx=1f(x) = \dfrac{x}{x} = 1.

For x < 0: f(x)=x−x=−1f(x) = \dfrac{x}{-x} = -1.

LHL:
lim⁡x→0−f(x)=−1\lim_{x\to 0^-}f(x) = -1

RHL:
lim⁡x→0+f(x)=1\lim_{x\to 0^+}f(x) = 1

Since LHL ≠\neq RHL, lim⁡x→0f(x)\lim_{x\to 0}f(x) does not exist.
27Find lim⁡x→5f(x)\lim_{x\to 5}f(x), where f(x)=∣x∣−5f(x) = |x| - 5Show solution
Given: f(x)=∣x∣−5f(x) = |x| - 5

For xx near 55 (both sides), x > 0, so ∣x∣=x|x| = x.

LHL:
lim⁡x→5−f(x)=lim⁡x→5−(x−5)=5−5=0\lim_{x\to 5^-}f(x) = \lim_{x\to 5^-}(x - 5) = 5 - 5 = 0

RHL:
lim⁡x→5+f(x)=lim⁡x→5+(x−5)=5−5=0\lim_{x\to 5^+}f(x) = \lim_{x\to 5^+}(x - 5) = 5 - 5 = 0

Since LHL == RHL =0= 0:
lim⁡x→5f(x)=0\lim_{x\to 5}f(x) = 0
28Suppose f(x)={a+bx,amp;xlt;14,amp;x=1b−ax,amp;xgt;1f(x) = \begin{cases} a + bx, & x < 1 \\ 4, & x = 1 \\ b - ax, & x > 1 \end{cases} and if lim⁡x→1f(x)=f(1)\lim_{x\to 1}f(x) = f(1), what are possible values of aa and bb?Show solution
Given: f(1)=4f(1) = 4 and lim⁡x→1f(x)=f(1)=4\lim_{x\to 1}f(x) = f(1) = 4.

For the limit to exist, LHL must equal RHL.

LHL:
lim⁡x→1−f(x)=lim⁡x→1−(a+bx)=a+b\lim_{x\to 1^-}f(x) = \lim_{x\to 1^-}(a + bx) = a + b

RHL:
lim⁡x→1+f(x)=lim⁡x→1+(b−ax)=b−a\lim_{x\to 1^+}f(x) = \lim_{x\to 1^+}(b - ax) = b - a

For the limit to exist:
a+b=b−a  ⟹  2a=0  ⟹  a=0a + b = b - a \implies 2a = 0 \implies a = 0

For the limit to equal f(1)=4f(1) = 4:
a+b=4  ⟹  0+b=4  ⟹  b=4a + b = 4 \implies 0 + b = 4 \implies b = 4

Answer: a=0a = 0 and b=4b = 4.
29Let a1,a2,…,ana_1, a_2, \ldots, a_n be fixed real numbers and define f(x)=(x−a1)(x−a2)⋯(x−an)f(x) = (x - a_1)(x - a_2)\cdots(x - a_n). What is lim⁡x→a1f(x)\lim_{x\to a_1}f(x)? For some a≠a1,a2,…,ana \neq a_1, a_2, \ldots, a_n, compute lim⁡x→af(x)\lim_{x\to a}f(x).Show solution
Given: f(x)=(x−a1)(x−a2)⋯(x−an)f(x) = (x - a_1)(x - a_2)\cdots(x - a_n)

Finding lim⁡x→a1f(x)\lim_{x\to a_1}f(x):

Since f(x)f(x) is a polynomial, the limit equals the value at x=a1x = a_1:
lim⁡x→a1f(x)=(a1−a1)(a1−a2)⋯(a1−an)=0\lim_{x\to a_1}f(x) = (a_1 - a_1)(a_1 - a_2)\cdots(a_1 - a_n) = 0

Finding lim⁡x→af(x)\lim_{x\to a}f(x) for a≠a1,a2,…,ana \neq a_1, a_2, \ldots, a_n:

Again by direct substitution:
lim⁡x→af(x)=(a−a1)(a−a2)⋯(a−an)\lim_{x\to a}f(x) = (a - a_1)(a - a_2)\cdots(a - a_n)

This is a non-zero finite value since aa is different from all aia_i.
30If f(x)={∣x∣+1,amp;xlt;00,amp;x=0∣x∣−1,amp;xgt;0f(x) = \begin{cases} |x| + 1, & x < 0 \\ 0, & x = 0 \\ |x| - 1, & x > 0 \end{cases}. For what value(s) of aa does lim⁡x→af(x)\lim_{x\to a}f(x) exist?Show solution
Given: f(x)={∣x∣+1,amp;xlt;00,amp;x=0∣x∣−1,amp;xgt;0f(x) = \begin{cases} |x| + 1, & x < 0 \\ 0, & x = 0 \\ |x| - 1, & x > 0 \end{cases}

Case 1: a=0a = 0

LHL: lim⁡x→0−f(x)=lim⁡x→0−(∣x∣+1)=0+1=1\lim_{x\to 0^-}f(x) = \lim_{x\to 0^-}(|x| + 1) = 0 + 1 = 1

RHL: lim⁡x→0+f(x)=lim⁡x→0+(∣x∣−1)=0−1=−1\lim_{x\to 0^+}f(x) = \lim_{x\to 0^+}(|x| - 1) = 0 - 1 = -1

LHL ≠\neq RHL, so the limit does not exist at a=0a = 0.

Case 2: a < 0

For xx near a < 0, f(x)=∣x∣+1=−x+1f(x) = |x| + 1 = -x + 1 (since x < 0).

lim⁡x→af(x)=−a+1\lim_{x\to a}f(x) = -a + 1

LHL == RHL, so the limit exists for all a < 0.

Case 3: a > 0

For xx near a > 0, f(x)=∣x∣−1=x−1f(x) = |x| - 1 = x - 1.

lim⁡x→af(x)=a−1\lim_{x\to a}f(x) = a - 1

LHL == RHL, so the limit exists for all a > 0.

Conclusion: lim⁡x→af(x)\lim_{x\to a}f(x) exists for all a≠0a \neq 0, i.e., for all a∈(−∞,0)∪(0,∞)a \in (-\infty, 0) \cup (0, \infty).
31If the function f(x)f(x) satisfies lim⁡x→1f(x)−2x2−1=π\lim_{x\to 1}\dfrac{f(x) - 2}{x^2 - 1} = \pi, evaluate lim⁡x→1f(x)\lim_{x\to 1}f(x).Show solution
Given: lim⁡x→1f(x)−2x2−1=π\lim_{x\to 1}\dfrac{f(x) - 2}{x^2 - 1} = \pi

Working:

As x→1x \to 1, the denominator x2−1→0x^2 - 1 \to 0. For the limit to be finite (equal to π\pi), the numerator must also →0\to 0.

Therefore:
lim⁡x→1[f(x)−2]=lim⁡x→1f(x)−2x2−1⋅lim⁡x→1(x2−1)=π×0=0\lim_{x\to 1}[f(x) - 2] = \lim_{x\to 1}\frac{f(x)-2}{x^2-1} \cdot \lim_{x\to 1}(x^2-1) = \pi \times 0 = 0

lim⁡x→1f(x)−2=0\lim_{x\to 1}f(x) - 2 = 0

lim⁡x→1f(x)=2\lim_{x\to 1}f(x) = 2

Answer: lim⁡x→1f(x)=2\lim_{x\to 1}f(x) = 2
32If f(x)={mx2+n,amp;xlt;0nx+m,amp;0≤x≤1nx3+m,amp;xgt;1f(x) = \begin{cases} mx^2 + n, & x < 0 \\ nx + m, & 0 \leq x \leq 1 \\ nx^3 + m, & x > 1 \end{cases}. For what integers mm and nn does both lim⁡x→0f(x)\lim_{x\to 0}f(x) and lim⁡x→1f(x)\lim_{x\to 1}f(x) exist?Show solution
Given: f(x)={mx2+n,amp;xlt;0nx+m,amp;0≤x≤1nx3+m,amp;xgt;1f(x) = \begin{cases} mx^2 + n, & x < 0 \\ nx + m, & 0 \leq x \leq 1 \\ nx^3 + m, & x > 1 \end{cases}

Condition for lim⁡x→0f(x)\lim_{x\to 0}f(x) to exist:

LHL: lim⁡x→0−f(x)=lim⁡x→0−(mx2+n)=n\lim_{x\to 0^-}f(x) = \lim_{x\to 0^-}(mx^2 + n) = n

RHL: lim⁡x→0+f(x)=lim⁡x→0+(nx+m)=m\lim_{x\to 0^+}f(x) = \lim_{x\to 0^+}(nx + m) = m

For limit to exist: LHL == RHL ⇒\Rightarrow n=mn = m.

Condition for lim⁡x→1f(x)\lim_{x\to 1}f(x) to exist:

LHL: lim⁡x→1−f(x)=lim⁡x→1−(nx+m)=n+m\lim_{x\to 1^-}f(x) = \lim_{x\to 1^-}(nx + m) = n + m

RHL: lim⁡x→1+f(x)=lim⁡x→1+(nx3+m)=n+m\lim_{x\to 1^+}f(x) = \lim_{x\to 1^+}(nx^3 + m) = n + m

LHL == RHL =n+m= n + m for all values of mm and nn. So lim⁡x→1f(x)\lim_{x\to 1}f(x) exists for all integers mm and nn.

Conclusion: Both limits exist when m=nm = n (where mm and nn are any equal integers).

Exercise 12.2

1Find the derivative of x2−2x^2 - 2 at x=10x = 10.Show solution
Given: f(x)=x2−2f(x) = x^2 - 2

Formula: f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h\to 0}\dfrac{f(x+h)-f(x)}{h}

Working:
f′(x)=lim⁡h→0(x+h)2−2−(x2−2)h=lim⁡h→0x2+2xh+h2−x2hf'(x) = \lim_{h\to 0}\frac{(x+h)^2 - 2 - (x^2 - 2)}{h} = \lim_{h\to 0}\frac{x^2 + 2xh + h^2 - x^2}{h}
=lim⁡h→02xh+h2h=lim⁡h→0(2x+h)=2x= \lim_{h\to 0}\frac{2xh + h^2}{h} = \lim_{h\to 0}(2x + h) = 2x

At x=10x = 10:
f′(10)=2(10)=20f'(10) = 2(10) = 20

Answer: 2020
2Find the derivative of xx at x=1x = 1.Show solution
Given: f(x)=xf(x) = x

Working:
f′(x)=lim⁡h→0(x+h)−xh=lim⁡h→0hh=1f'(x) = \lim_{h\to 0}\frac{(x+h) - x}{h} = \lim_{h\to 0}\frac{h}{h} = 1

At x=1x = 1: f′(1)=1f'(1) = 1

Answer: 11
3Find the derivative of 99x99x at x=100x = 100.Show solution
Given: f(x)=99xf(x) = 99x

Working:
f′(x)=lim⁡h→099(x+h)−99xh=lim⁡h→099hh=99f'(x) = \lim_{h\to 0}\frac{99(x+h) - 99x}{h} = \lim_{h\to 0}\frac{99h}{h} = 99

At x=100x = 100: f′(100)=99f'(100) = 99

Answer: 9999
4Find the derivative of the following functions from first principle.
(i) x3−27x^3 - 27
(ii) (x−1)(x−2)(x-1)(x-2)
(iii) 1x2\dfrac{1}{x^2}
(iv) x+1x−1\dfrac{x+1}{x-1}Show solution
(i) f(x)=x3−27f(x) = x^3 - 27

f′(x)=lim⁡h→0(x+h)3−27−(x3−27)hf'(x) = \lim_{h\to 0}\frac{(x+h)^3 - 27 - (x^3 - 27)}{h}
=lim⁡h→0x3+3x2h+3xh2+h3−x3h= \lim_{h\to 0}\frac{x^3 + 3x^2h + 3xh^2 + h^3 - x^3}{h}
=lim⁡h→0(3x2+3xh+h2)=3x2= \lim_{h\to 0}(3x^2 + 3xh + h^2) = 3x^2

Answer: f′(x)=3x2f'(x) = 3x^2

---

(ii) f(x)=(x−1)(x−2)=x2−3x+2f(x) = (x-1)(x-2) = x^2 - 3x + 2

f′(x)=lim⁡h→0(x+h)2−3(x+h)+2−(x2−3x+2)hf'(x) = \lim_{h\to 0}\frac{(x+h)^2 - 3(x+h) + 2 - (x^2 - 3x + 2)}{h}
=lim⁡h→02xh+h2−3hh=lim⁡h→0(2x+h−3)=2x−3= \lim_{h\to 0}\frac{2xh + h^2 - 3h}{h} = \lim_{h\to 0}(2x + h - 3) = 2x - 3

Answer: f′(x)=2x−3f'(x) = 2x - 3

---

(iii) f(x)=1x2f(x) = \dfrac{1}{x^2}

f′(x)=lim⁡h→01(x+h)2−1x2h=lim⁡h→0x2−(x+h)2h⋅x2(x+h)2f'(x) = \lim_{h\to 0}\frac{\dfrac{1}{(x+h)^2} - \dfrac{1}{x^2}}{h} = \lim_{h\to 0}\frac{x^2 - (x+h)^2}{h \cdot x^2(x+h)^2}
=lim⁡h→0−2xh−h2h⋅x2(x+h)2=lim⁡h→0−(2x+h)x2(x+h)2=−2xx4=−2x3= \lim_{h\to 0}\frac{-2xh - h^2}{h \cdot x^2(x+h)^2} = \lim_{h\to 0}\frac{-(2x + h)}{x^2(x+h)^2} = \frac{-2x}{x^4} = \frac{-2}{x^3}

Answer: f′(x)=−2x3f'(x) = -\dfrac{2}{x^3}

---

(iv) f(x)=x+1x−1f(x) = \dfrac{x+1}{x-1}

f′(x)=lim⁡h→0x+h+1x+h−1−x+1x−1hf'(x) = \lim_{h\to 0}\frac{\dfrac{x+h+1}{x+h-1} - \dfrac{x+1}{x-1}}{h}
=lim⁡h→0(x+h+1)(x−1)−(x+1)(x+h−1)h(x+h−1)(x−1)= \lim_{h\to 0}\frac{(x+h+1)(x-1) - (x+1)(x+h-1)}{h(x+h-1)(x-1)}

Expanding numerator:
(x+h+1)(x−1)=x2−x+hx−h+x−1=x2+hx−h−1(x+h+1)(x-1) = x^2 - x + hx - h + x - 1 = x^2 + hx - h - 1
(x+1)(x+h−1)=x2+xh−x+x+h−1=x2+xh+h−1(x+1)(x+h-1) = x^2 + xh - x + x + h - 1 = x^2 + xh + h - 1

Numerator =(x2+hx−h−1)−(x2+xh+h−1)=−2h= (x^2 + hx - h - 1) - (x^2 + xh + h - 1) = -2h

f′(x)=lim⁡h→0−2hh(x+h−1)(x−1)=lim⁡h→0−2(x+h−1)(x−1)=−2(x−1)2f'(x) = \lim_{h\to 0}\frac{-2h}{h(x+h-1)(x-1)} = \lim_{h\to 0}\frac{-2}{(x+h-1)(x-1)} = \frac{-2}{(x-1)^2}

Answer: f′(x)=−2(x−1)2f'(x) = \dfrac{-2}{(x-1)^2}
5For the function f(x)=x100100+x9999+⋯+x22+x+1f(x) = \dfrac{x^{100}}{100} + \dfrac{x^{99}}{99} + \cdots + \dfrac{x^2}{2} + x + 1. Prove that f′(1)=100f′(0)f'(1) = 100f'(0).Show solution
Given: f(x)=x100100+x9999+⋯+x22+x+1f(x) = \dfrac{x^{100}}{100} + \dfrac{x^{99}}{99} + \cdots + \dfrac{x^2}{2} + x + 1

Finding f′(x)f'(x):

Using ddx(xn)=nxn−1\dfrac{d}{dx}(x^n) = nx^{n-1}:
f′(x)=100x99100+99x9899+⋯+2x2+1f'(x) = \frac{100x^{99}}{100} + \frac{99x^{98}}{99} + \cdots + \frac{2x}{2} + 1
f′(x)=x99+x98+⋯+x+1f'(x) = x^{99} + x^{98} + \cdots + x + 1

Finding f′(1)f'(1):
f′(1)=199+198+⋯+1+1=1+1+⋯+1⏟100 terms=100f'(1) = 1^{99} + 1^{98} + \cdots + 1 + 1 = \underbrace{1 + 1 + \cdots + 1}_{100 \text{ terms}} = 100

Finding f′(0)f'(0):
f′(0)=0+0+⋯+0+1=1f'(0) = 0 + 0 + \cdots + 0 + 1 = 1

Verification:
100⋅f′(0)=100×1=100=f′(1)100 \cdot f'(0) = 100 \times 1 = 100 = f'(1)

Hence, f′(1)=100f′(0)f'(1) = 100f'(0). ■\hspace{2cm}\blacksquare
6Find the derivative of xn+axn−1+a2xn−2+…+an−1x+anx^n + ax^{n-1} + a^2x^{n-2} + \ldots + a^{n-1}x + a^n for some fixed real number aa.Show solution
Given: f(x)=xn+axn−1+a2xn−2+⋯+an−1x+anf(x) = x^n + ax^{n-1} + a^2x^{n-2} + \cdots + a^{n-1}x + a^n

Using ddx(xk)=kxk−1\dfrac{d}{dx}(x^k) = kx^{k-1} and ddx(constant)=0\dfrac{d}{dx}(\text{constant}) = 0:

f′(x)=nxn−1+a(n−1)xn−2+a2(n−2)xn−3+⋯+an−1f'(x) = nx^{n-1} + a(n-1)x^{n-2} + a^2(n-2)x^{n-3} + \cdots + a^{n-1}

f′(x)=nxn−1+(n−1)axn−2+(n−2)a2xn−3+⋯+an−1\boxed{f'(x) = nx^{n-1} + (n-1)ax^{n-2} + (n-2)a^2x^{n-3} + \cdots + a^{n-1}}
7For some constants aa and bb, find the derivative of:
(i) (x−a)(x−b)(x-a)(x-b)
(ii) (ax2+b)2(ax^2+b)^2
(iii) x−ax−b\dfrac{x-a}{x-b}Show solution
(i) f(x)=(x−a)(x−b)=x2−(a+b)x+abf(x) = (x-a)(x-b) = x^2 - (a+b)x + ab

f′(x)=2x−(a+b)f'(x) = 2x - (a+b)

---

(ii) f(x)=(ax2+b)2=a2x4+2abx2+b2f(x) = (ax^2 + b)^2 = a^2x^4 + 2abx^2 + b^2

f′(x)=4a2x3+4abx=4ax(ax2+b)f'(x) = 4a^2x^3 + 4abx = 4ax(ax^2 + b)

---

(iii) f(x)=x−ax−bf(x) = \dfrac{x-a}{x-b}

Using the quotient rule (uv)′=u′v−uv′v2\left(\dfrac{u}{v}\right)' = \dfrac{u'v - uv'}{v^2}:

u=x−a⇒u′=1u = x - a \Rightarrow u' = 1; v=x−b⇒v′=1v = x - b \Rightarrow v' = 1

f′(x)=(1)(x−b)−(x−a)(1)(x−b)2=x−b−x+a(x−b)2=a−b(x−b)2f'(x) = \frac{(1)(x-b) - (x-a)(1)}{(x-b)^2} = \frac{x - b - x + a}{(x-b)^2} = \frac{a - b}{(x-b)^2}
8Find the derivative of xn−anx−a\dfrac{x^n - a^n}{x - a} for some constant aa.Show solution
Given: f(x)=xn−anx−af(x) = \dfrac{x^n - a^n}{x - a}

Using the quotient rule:

u=xn−an⇒u′=nxn−1u = x^n - a^n \Rightarrow u' = nx^{n-1}; v=x−a⇒v′=1v = x - a \Rightarrow v' = 1

f′(x)=nxn−1(x−a)−(xn−an)(1)(x−a)2f'(x) = \frac{nx^{n-1}(x-a) - (x^n - a^n)(1)}{(x-a)^2}

=nxn−naxn−1−xn+an(x−a)2= \frac{nx^n - nax^{n-1} - x^n + a^n}{(x-a)^2}

=(n−1)xn−naxn−1+an(x−a)2= \frac{(n-1)x^n - nax^{n-1} + a^n}{(x-a)^2}
9Find the derivative of:
(i) 2x−342x - \dfrac{3}{4}
(ii) (5x3+3x−1)(x−1)(5x^3 + 3x - 1)(x-1)
(iii) x−3(5+3x)x^{-3}(5 + 3x)
(iv) x5(3−6x−9)x^5(3 - 6x^{-9})
(v) x−4(3−4x−5)x^{-4}(3 - 4x^{-5})
(vi) 2x+1−x23x−1\dfrac{2}{x+1} - \dfrac{x^2}{3x-1}Show solution
(i) f(x)=2x−34f(x) = 2x - \dfrac{3}{4}

f′(x)=2f'(x) = 2

---

(ii) f(x)=(5x3+3x−1)(x−1)f(x) = (5x^3 + 3x - 1)(x - 1)

Using product rule: (uv)′=u′v+uv′(uv)' = u'v + uv'

u=5x3+3x−1, u′=15x2+3u = 5x^3 + 3x - 1,\ u' = 15x^2 + 3
v=x−1, v′=1v = x - 1,\ v' = 1

f′(x)=(15x2+3)(x−1)+(5x3+3x−1)(1)f'(x) = (15x^2 + 3)(x-1) + (5x^3 + 3x - 1)(1)
=15x3−15x2+3x−3+5x3+3x−1= 15x^3 - 15x^2 + 3x - 3 + 5x^3 + 3x - 1
=20x3−15x2+6x−4= 20x^3 - 15x^2 + 6x - 4

---

(iii) f(x)=x−3(5+3x)=5x−3+3x−2f(x) = x^{-3}(5 + 3x) = 5x^{-3} + 3x^{-2}

f′(x)=−15x−4−6x−3=−15x4−6x3f'(x) = -15x^{-4} - 6x^{-3} = -\frac{15}{x^4} - \frac{6}{x^3}

---

(iv) f(x)=x5(3−6x−9)=3x5−6x−4f(x) = x^5(3 - 6x^{-9}) = 3x^5 - 6x^{-4}

f′(x)=15x4+24x−5=15x4+24x5f'(x) = 15x^4 + 24x^{-5} = 15x^4 + \frac{24}{x^5}

---

(v) f(x)=x−4(3−4x−5)=3x−4−4x−9f(x) = x^{-4}(3 - 4x^{-5}) = 3x^{-4} - 4x^{-9}

f′(x)=−12x−5+36x−10=−12x5+36x10f'(x) = -12x^{-5} + 36x^{-10} = -\frac{12}{x^5} + \frac{36}{x^{10}}

---

(vi) f(x)=2x+1−x23x−1f(x) = \dfrac{2}{x+1} - \dfrac{x^2}{3x-1}

Differentiate each term using quotient rule.

For 2x+1\dfrac{2}{x+1}:
ddx(2x+1)=0⋅(x+1)−2⋅1(x+1)2=−2(x+1)2\frac{d}{dx}\left(\frac{2}{x+1}\right) = \frac{0 \cdot (x+1) - 2 \cdot 1}{(x+1)^2} = \frac{-2}{(x+1)^2}

For x23x−1\dfrac{x^2}{3x-1}:
ddx(x23x−1)=2x(3x−1)−x2⋅3(3x−1)2=6x2−2x−3x2(3x−1)2=3x2−2x(3x−1)2\frac{d}{dx}\left(\frac{x^2}{3x-1}\right) = \frac{2x(3x-1) - x^2 \cdot 3}{(3x-1)^2} = \frac{6x^2 - 2x - 3x^2}{(3x-1)^2} = \frac{3x^2 - 2x}{(3x-1)^2}

f′(x)=−2(x+1)2−3x2−2x(3x−1)2f'(x) = \frac{-2}{(x+1)^2} - \frac{3x^2 - 2x}{(3x-1)^2}
10Find the derivative of cos⁡x\cos x from first principle.Show solution
Given: f(x)=cos⁡xf(x) = \cos x

Using first principle:
f′(x)=lim⁡h→0cos⁡(x+h)−cos⁡xhf'(x) = \lim_{h\to 0}\frac{\cos(x+h) - \cos x}{h}

Using cos⁡A−cos⁡B=−2sin⁡A+B2sin⁡A−B2\cos A - \cos B = -2\sin\dfrac{A+B}{2}\sin\dfrac{A-B}{2}:

cos⁡(x+h)−cos⁡x=−2sin⁡(x+h2)sin⁡(h2)\cos(x+h) - \cos x = -2\sin\left(x + \frac{h}{2}\right)\sin\left(\frac{h}{2}\right)

f′(x)=lim⁡h→0−2sin⁡(x+h2)sin⁡(h2)hf'(x) = \lim_{h\to 0}\frac{-2\sin\left(x + \frac{h}{2}\right)\sin\left(\frac{h}{2}\right)}{h}

=lim⁡h→0[−sin⁡(x+h2)⋅sin⁡(h/2)h/2]= \lim_{h\to 0}\left[-\sin\left(x + \frac{h}{2}\right) \cdot \frac{\sin(h/2)}{h/2}\right]

=−sin⁡(x+0)⋅1=−sin⁡x= -\sin(x + 0) \cdot 1 = -\sin x

Answer: ddx(cos⁡x)=−sin⁡x\dfrac{d}{dx}(\cos x) = -\sin x
11Find the derivative of the following functions:
(i) sin⁡xcos⁡x\sin x\cos x
(ii) sec⁡x\sec x
(iii) 5sec⁡x+4cos⁡x5\sec x + 4\cos x
(iv) csc⁡x\csc x
(v) 3cot⁡x+5csc⁡x3\cot x + 5\csc x
(vi) 5sin⁡x−6cos⁡x+75\sin x - 6\cos x + 7
(vii) 2tan⁡x−7sec⁡x2\tan x - 7\sec xShow solution
(i) f(x)=sin⁡xcos⁡xf(x) = \sin x \cos x

Using product rule:
f′(x)=cos⁡x⋅cos⁡x+sin⁡x⋅(−sin⁡x)=cos⁡2x−sin⁡2x=cos⁡2xf'(x) = \cos x \cdot \cos x + \sin x \cdot (-\sin x) = \cos^2 x - \sin^2 x = \cos 2x

---

(ii) f(x)=sec⁡x=1cos⁡xf(x) = \sec x = \dfrac{1}{\cos x}

Using quotient rule:
f′(x)=0⋅cos⁡x−1⋅(−sin⁡x)cos⁡2x=sin⁡xcos⁡2x=sec⁡xtan⁡xf'(x) = \frac{0 \cdot \cos x - 1 \cdot (-\sin x)}{\cos^2 x} = \frac{\sin x}{\cos^2 x} = \sec x \tan x

---

(iii) f(x)=5sec⁡x+4cos⁡xf(x) = 5\sec x + 4\cos x

f′(x)=5sec⁡xtan⁡x+4(−sin⁡x)=5sec⁡xtan⁡x−4sin⁡xf'(x) = 5\sec x\tan x + 4(-\sin x) = 5\sec x\tan x - 4\sin x

---

(iv) f(x)=csc⁡x=1sin⁡xf(x) = \csc x = \dfrac{1}{\sin x}

Using quotient rule:
f′(x)=0⋅sin⁡x−1⋅cos⁡xsin⁡2x=−cos⁡xsin⁡2x=−csc⁡xcot⁡xf'(x) = \frac{0 \cdot \sin x - 1 \cdot \cos x}{\sin^2 x} = \frac{-\cos x}{\sin^2 x} = -\csc x \cot x

---

(v) f(x)=3cot⁡x+5csc⁡xf(x) = 3\cot x + 5\csc x

Using ddx(cot⁡x)=−csc⁡2x\dfrac{d}{dx}(\cot x) = -\csc^2 x and ddx(csc⁡x)=−csc⁡xcot⁡x\dfrac{d}{dx}(\csc x) = -\csc x\cot x:
f′(x)=3(−csc⁡2x)+5(−csc⁡xcot⁡x)=−3csc⁡2x−5csc⁡xcot⁡xf'(x) = 3(-\csc^2 x) + 5(-\csc x\cot x) = -3\csc^2 x - 5\csc x\cot x

---

(vi) f(x)=5sin⁡x−6cos⁡x+7f(x) = 5\sin x - 6\cos x + 7

f′(x)=5cos⁡x−6(−sin⁡x)+0=5cos⁡x+6sin⁡xf'(x) = 5\cos x - 6(-\sin x) + 0 = 5\cos x + 6\sin x

---

(vii) f(x)=2tan⁡x−7sec⁡xf(x) = 2\tan x - 7\sec x

Using ddx(tan⁡x)=sec⁡2x\dfrac{d}{dx}(\tan x) = \sec^2 x and ddx(sec⁡x)=sec⁡xtan⁡x\dfrac{d}{dx}(\sec x) = \sec x\tan x:
f′(x)=2sec⁡2x−7sec⁡xtan⁡xf'(x) = 2\sec^2 x - 7\sec x\tan x

Miscellaneous Exercise on Chapter 12

1Find the derivative of the following functions from first principle:
(i) −x-x
(ii) (−x)−1(-x)^{-1}
(iii) sin⁡(x+1)\sin(x+1)
(iv) cos⁡(x−π8)\cos\left(x - \dfrac{\pi}{8}\right)Show solution
(i) f(x)=−xf(x) = -x

f′(x)=lim⁡h→0−(x+h)−(−x)h=lim⁡h→0−hh=−1f'(x) = \lim_{h\to 0}\frac{-(x+h)-(-x)}{h} = \lim_{h\to 0}\frac{-h}{h} = -1

Answer: f′(x)=−1f'(x) = -1

---

(ii) f(x)=(−x)−1=−1xf(x) = (-x)^{-1} = -\dfrac{1}{x}

f′(x)=lim⁡h→0−1x+h−(−1x)h=lim⁡h→01x−1x+hhf'(x) = \lim_{h\to 0}\frac{-\dfrac{1}{x+h} - \left(-\dfrac{1}{x}\right)}{h} = \lim_{h\to 0}\frac{\dfrac{1}{x} - \dfrac{1}{x+h}}{h}
=lim⁡h→0hx(x+h)h=lim⁡h→01x(x+h)=1x2= \lim_{h\to 0}\frac{\dfrac{h}{x(x+h)}}{h} = \lim_{h\to 0}\frac{1}{x(x+h)} = \frac{1}{x^2}

Answer: f′(x)=1x2f'(x) = \dfrac{1}{x^2}

---

(iii) f(x)=sin⁡(x+1)f(x) = \sin(x+1)

f′(x)=lim⁡h→0sin⁡(x+h+1)−sin⁡(x+1)hf'(x) = \lim_{h\to 0}\frac{\sin(x+h+1) - \sin(x+1)}{h}

Using sin⁡A−sin⁡B=2cos⁡A+B2sin⁡A−B2\sin A - \sin B = 2\cos\dfrac{A+B}{2}\sin\dfrac{A-B}{2}:

=lim⁡h→02cos⁡(x+1+h2)sin⁡(h2)h= \lim_{h\to 0}\frac{2\cos\left(x+1+\dfrac{h}{2}\right)\sin\left(\dfrac{h}{2}\right)}{h}
=lim⁡h→0cos⁡(x+1+h2)⋅sin⁡(h/2)h/2=cos⁡(x+1)⋅1=cos⁡(x+1)= \lim_{h\to 0}\cos\left(x+1+\frac{h}{2}\right)\cdot\frac{\sin(h/2)}{h/2} = \cos(x+1) \cdot 1 = \cos(x+1)

Answer: f′(x)=cos⁡(x+1)f'(x) = \cos(x+1)

---

(iv) f(x)=cos⁡(x−π8)f(x) = \cos\left(x - \dfrac{\pi}{8}\right)

f′(x)=lim⁡h→0cos⁡(x+h−π8)−cos⁡(x−π8)hf'(x) = \lim_{h\to 0}\frac{\cos\left(x+h-\dfrac{\pi}{8}\right) - \cos\left(x-\dfrac{\pi}{8}\right)}{h}

Using cos⁡A−cos⁡B=−2sin⁡A+B2sin⁡A−B2\cos A - \cos B = -2\sin\dfrac{A+B}{2}\sin\dfrac{A-B}{2}:

=lim⁡h→0−2sin⁡(x−π8+h2)sin⁡(h2)h= \lim_{h\to 0}\frac{-2\sin\left(x - \dfrac{\pi}{8} + \dfrac{h}{2}\right)\sin\left(\dfrac{h}{2}\right)}{h}
=−sin⁡(x−π8)⋅1=−sin⁡(x−π8)= -\sin\left(x - \frac{\pi}{8}\right) \cdot 1 = -\sin\left(x - \frac{\pi}{8}\right)

Answer: f′(x)=−sin⁡(x−π8)f'(x) = -\sin\left(x - \dfrac{\pi}{8}\right)
2Find the derivative of (x+a)(x + a).Show solution
Given: f(x)=x+af(x) = x + a

f′(x)=ddx(x)+ddx(a)=1+0=1f'(x) = \frac{d}{dx}(x) + \frac{d}{dx}(a) = 1 + 0 = 1

Answer: 11
3Find the derivative of (px+q)(rx+s)(px + q)\left(\dfrac{r}{x} + s\right).Show solution
Given: f(x)=(px+q)(rx+s)=pr+psx+qrx+qsf(x) = (px + q)\left(\dfrac{r}{x} + s\right) = pr + psx + \dfrac{qr}{x} + qs

f′(x)=ps−qrx2f'(x) = ps - \frac{qr}{x^2}

Alternatively using product rule:
u=px+q, u′=pu = px + q,\ u' = p; v=rx+s, v′=−rx2v = \dfrac{r}{x} + s,\ v' = -\dfrac{r}{x^2}

f′(x)=p(rx+s)+(px+q)(−rx2)=prx+ps−prx−qrx2=ps−qrx2f'(x) = p\left(\frac{r}{x} + s\right) + (px + q)\left(-\frac{r}{x^2}\right) = \frac{pr}{x} + ps - \frac{pr}{x} - \frac{qr}{x^2} = ps - \frac{qr}{x^2}

Answer: f′(x)=ps−qrx2f'(x) = ps - \dfrac{qr}{x^2}
4Find the derivative of (ax+b)(cx+d)2(ax + b)(cx + d)^2.Show solution
Given: f(x)=(ax+b)(cx+d)2f(x) = (ax + b)(cx + d)^2

Using product rule: u=ax+b, u′=au = ax + b,\ u' = a; v=(cx+d)2, v′=2c(cx+d)v = (cx+d)^2,\ v' = 2c(cx+d)

f′(x)=a(cx+d)2+(ax+b)⋅2c(cx+d)f'(x) = a(cx+d)^2 + (ax+b) \cdot 2c(cx+d)
=(cx+d)[a(cx+d)+2c(ax+b)]= (cx+d)\left[a(cx+d) + 2c(ax+b)\right]
=(cx+d)(acx+ad+2acx+2bc)= (cx+d)(acx + ad + 2acx + 2bc)
=(cx+d)(3acx+ad+2bc)= (cx+d)(3acx + ad + 2bc)

Answer: f′(x)=(cx+d)(3acx+ad+2bc)f'(x) = (cx+d)(3acx + ad + 2bc)
5Find the derivative of ax+bcx+d\dfrac{ax + b}{cx + d}.Show solution
Given: f(x)=ax+bcx+df(x) = \dfrac{ax+b}{cx+d}

Using quotient rule: u=ax+b, u′=au = ax+b,\ u' = a; v=cx+d, v′=cv = cx+d,\ v' = c

f′(x)=a(cx+d)−(ax+b)c(cx+d)2=acx+ad−acx−bc(cx+d)2=ad−bc(cx+d)2f'(x) = \frac{a(cx+d) - (ax+b)c}{(cx+d)^2} = \frac{acx + ad - acx - bc}{(cx+d)^2} = \frac{ad - bc}{(cx+d)^2}

Answer: f′(x)=ad−bc(cx+d)2f'(x) = \dfrac{ad - bc}{(cx+d)^2}
6Find the derivative of 1+1x1−1x\dfrac{1 + \dfrac{1}{x}}{1 - \dfrac{1}{x}}.Show solution
Given: f(x)=1+1x1−1x=x+1xx−1x=x+1x−1f(x) = \dfrac{1 + \dfrac{1}{x}}{1 - \dfrac{1}{x}} = \dfrac{\dfrac{x+1}{x}}{\dfrac{x-1}{x}} = \dfrac{x+1}{x-1}

Using quotient rule: u=x+1, u′=1u = x+1,\ u' = 1; v=x−1, v′=1v = x-1,\ v' = 1

f′(x)=1⋅(x−1)−(x+1)⋅1(x−1)2=x−1−x−1(x−1)2=−2(x−1)2f'(x) = \frac{1 \cdot (x-1) - (x+1) \cdot 1}{(x-1)^2} = \frac{x-1-x-1}{(x-1)^2} = \frac{-2}{(x-1)^2}

Answer: f′(x)=−2(x−1)2f'(x) = \dfrac{-2}{(x-1)^2}
7Find the derivative of 1ax2+bx+c\dfrac{1}{ax^2 + bx + c}.Show solution
Given: f(x)=1ax2+bx+cf(x) = \dfrac{1}{ax^2 + bx + c}

Using quotient rule (or chain rule): u=1, u′=0u = 1,\ u' = 0; v=ax2+bx+c, v′=2ax+bv = ax^2+bx+c,\ v' = 2ax+b

f′(x)=0⋅(ax2+bx+c)−1⋅(2ax+b)(ax2+bx+c)2=−(2ax+b)(ax2+bx+c)2f'(x) = \frac{0 \cdot (ax^2+bx+c) - 1 \cdot (2ax+b)}{(ax^2+bx+c)^2} = \frac{-(2ax+b)}{(ax^2+bx+c)^2}

Answer: f′(x)=−(2ax+b)(ax2+bx+c)2f'(x) = \dfrac{-(2ax+b)}{(ax^2+bx+c)^2}
8Find the derivative of ax+bpx2+qx+r\dfrac{ax + b}{px^2 + qx + r}.Show solution
Given: f(x)=ax+bpx2+qx+rf(x) = \dfrac{ax+b}{px^2+qx+r}

u=ax+b, u′=au = ax+b,\ u' = a; v=px2+qx+r, v′=2px+qv = px^2+qx+r,\ v' = 2px+q

f′(x)=a(px2+qx+r)−(ax+b)(2px+q)(px2+qx+r)2f'(x) = \frac{a(px^2+qx+r) - (ax+b)(2px+q)}{(px^2+qx+r)^2}

Expanding numerator:
=apx2+aqx+ar−(2apx2+aqx+2bpx+bq)= apx^2 + aqx + ar - (2apx^2 + aqx + 2bpx + bq)
=apx2+aqx+ar−2apx2−aqx−2bpx−bq= apx^2 + aqx + ar - 2apx^2 - aqx - 2bpx - bq
=−apx2+ar−2bpx−bq= -apx^2 + ar - 2bpx - bq
=−(apx2+2bpx+bq−ar)= -(apx^2 + 2bpx + bq - ar)

f′(x)=−(apx2+2bpx+bq−ar)(px2+qx+r)2f'(x) = \frac{-(apx^2 + 2bpx + bq - ar)}{(px^2+qx+r)^2}
9Find the derivative of px2+qx+rax+b\dfrac{px^2 + qx + r}{ax + b}.Show solution
Given: f(x)=px2+qx+rax+bf(x) = \dfrac{px^2+qx+r}{ax+b}

u=px2+qx+r, u′=2px+qu = px^2+qx+r,\ u' = 2px+q; v=ax+b, v′=av = ax+b,\ v' = a

f′(x)=(2px+q)(ax+b)−(px2+qx+r)⋅a(ax+b)2f'(x) = \frac{(2px+q)(ax+b) - (px^2+qx+r) \cdot a}{(ax+b)^2}

Expanding numerator:
=2apx2+2bpx+aqx+bq−apx2−aqx−ar= 2apx^2 + 2bpx + aqx + bq - apx^2 - aqx - ar
=apx2+2bpx+bq−ar= apx^2 + 2bpx + bq - ar

f′(x)=apx2+2bpx+bq−ar(ax+b)2f'(x) = \frac{apx^2 + 2bpx + bq - ar}{(ax+b)^2}
10Find the derivative of ax4−bx2+cos⁡x\dfrac{a}{x^4} - \dfrac{b}{x^2} + \cos x.Show solution
Given: f(x)=ax−4−bx−2+cos⁡xf(x) = ax^{-4} - bx^{-2} + \cos x

f′(x)=−4ax−5+2bx−3−sin⁡x=−4ax5+2bx3−sin⁡xf'(x) = -4ax^{-5} + 2bx^{-3} - \sin x = -\frac{4a}{x^5} + \frac{2b}{x^3} - \sin x
11Find the derivative of 4x−24\sqrt{x} - 2.Show solution
Given: f(x)=4x1/2−2f(x) = 4x^{1/2} - 2

f′(x)=4⋅12x−1/2−0=2xf'(x) = 4 \cdot \frac{1}{2}x^{-1/2} - 0 = \frac{2}{\sqrt{x}}

Answer: f′(x)=2xf'(x) = \dfrac{2}{\sqrt{x}}
12Find the derivative of (ax+b)n(ax + b)^n.Show solution
Given: f(x)=(ax+b)nf(x) = (ax+b)^n

Using chain rule (or first principles with binomial theorem):

f′(x)=n(ax+b)n−1⋅a=na(ax+b)n−1f'(x) = n(ax+b)^{n-1} \cdot a = na(ax+b)^{n-1}

Answer: f′(x)=na(ax+b)n−1f'(x) = na(ax+b)^{n-1}
13Find the derivative of (ax+b)n(cx+d)m(ax + b)^n(cx + d)^m.Show solution
Given: f(x)=(ax+b)n(cx+d)mf(x) = (ax+b)^n(cx+d)^m

Using product rule: u=(ax+b)n, u′=na(ax+b)n−1u = (ax+b)^n,\ u' = na(ax+b)^{n-1}; v=(cx+d)m, v′=mc(cx+d)m−1v = (cx+d)^m,\ v' = mc(cx+d)^{m-1}

f′(x)=na(ax+b)n−1(cx+d)m+(ax+b)n⋅mc(cx+d)m−1f'(x) = na(ax+b)^{n-1}(cx+d)^m + (ax+b)^n \cdot mc(cx+d)^{m-1}
=(ax+b)n−1(cx+d)m−1[na(cx+d)+mc(ax+b)]= (ax+b)^{n-1}(cx+d)^{m-1}\left[na(cx+d) + mc(ax+b)\right]
14Find the derivative of sin⁡(x+a)\sin(x + a).Show solution
Given: f(x)=sin⁡(x+a)f(x) = \sin(x+a)

Using first principle:
f′(x)=lim⁡h→0sin⁡(x+h+a)−sin⁡(x+a)hf'(x) = \lim_{h\to 0}\frac{\sin(x+h+a) - \sin(x+a)}{h}

Using sin⁡A−sin⁡B=2cos⁡A+B2sin⁡A−B2\sin A - \sin B = 2\cos\dfrac{A+B}{2}\sin\dfrac{A-B}{2}:

=lim⁡h→02cos⁡(x+a+h2)sin⁡(h2)h=cos⁡(x+a)⋅1=cos⁡(x+a)= \lim_{h\to 0}\frac{2\cos\left(x+a+\dfrac{h}{2}\right)\sin\left(\dfrac{h}{2}\right)}{h} = \cos(x+a) \cdot 1 = \cos(x+a)

Answer: f′(x)=cos⁡(x+a)f'(x) = \cos(x+a)
15Find the derivative of csc⁡xcot⁡x\csc x \cot x.Show solution
Given: f(x)=csc⁡xcot⁡xf(x) = \csc x \cot x

Using product rule: u=csc⁡x, u′=−csc⁡xcot⁡xu = \csc x,\ u' = -\csc x\cot x; v=cot⁡x, v′=−csc⁡2xv = \cot x,\ v' = -\csc^2 x

f′(x)=(−csc⁡xcot⁡x)(cot⁡x)+csc⁡x(−csc⁡2x)f'(x) = (-\csc x\cot x)(\cot x) + \csc x(-\csc^2 x)
=−csc⁡xcot⁡2x−csc⁡3x= -\csc x\cot^2 x - \csc^3 x
=−csc⁡x(cot⁡2x+csc⁡2x)= -\csc x(\cot^2 x + \csc^2 x)
16Find the derivative of cos⁡x1+sin⁡x\dfrac{\cos x}{1 + \sin x}.Show solution
Given: f(x)=cos⁡x1+sin⁡xf(x) = \dfrac{\cos x}{1 + \sin x}

u=cos⁡x, u′=−sin⁡xu = \cos x,\ u' = -\sin x; v=1+sin⁡x, v′=cos⁡xv = 1 + \sin x,\ v' = \cos x

f′(x)=−sin⁡x(1+sin⁡x)−cos⁡x⋅cos⁡x(1+sin⁡x)2f'(x) = \frac{-\sin x(1+\sin x) - \cos x \cdot \cos x}{(1+\sin x)^2}
=−sin⁡x−sin⁡2x−cos⁡2x(1+sin⁡x)2= \frac{-\sin x - \sin^2 x - \cos^2 x}{(1+\sin x)^2}
=−sin⁡x−(sin⁡2x+cos⁡2x)(1+sin⁡x)2= \frac{-\sin x - (\sin^2 x + \cos^2 x)}{(1+\sin x)^2}
=−sin⁡x−1(1+sin⁡x)2=−(1+sin⁡x)(1+sin⁡x)2=−11+sin⁡x= \frac{-\sin x - 1}{(1+\sin x)^2} = \frac{-(1+\sin x)}{(1+\sin x)^2} = \frac{-1}{1+\sin x}
17Find the derivative of sin⁡x+cos⁡xsin⁡x−cos⁡x\dfrac{\sin x + \cos x}{\sin x - \cos x}.Show solution
Given: f(x)=sin⁡x+cos⁡xsin⁡x−cos⁡xf(x) = \dfrac{\sin x + \cos x}{\sin x - \cos x}

u=sin⁡x+cos⁡x, u′=cos⁡x−sin⁡xu = \sin x + \cos x,\ u' = \cos x - \sin x
v=sin⁡x−cos⁡x, v′=cos⁡x+sin⁡xv = \sin x - \cos x,\ v' = \cos x + \sin x

f′(x)=(cos⁡x−sin⁡x)(sin⁡x−cos⁡x)−(sin⁡x+cos⁡x)(cos⁡x+sin⁡x)(sin⁡x−cos⁡x)2f'(x) = \frac{(\cos x - \sin x)(\sin x - \cos x) - (\sin x + \cos x)(\cos x + \sin x)}{(\sin x - \cos x)^2}

Numerator:
=−(cos⁡x−sin⁡x)2−(sin⁡x+cos⁡x)2= -(\cos x - \sin x)^2 - (\sin x + \cos x)^2
=−(cos⁡2x−2sin⁡xcos⁡x+sin⁡2x)−(sin⁡2x+2sin⁡xcos⁡x+cos⁡2x)= -(\cos^2 x - 2\sin x\cos x + \sin^2 x) - (\sin^2 x + 2\sin x\cos x + \cos^2 x)
=−(1−2sin⁡xcos⁡x)−(1+2sin⁡xcos⁡x)= -(1 - 2\sin x\cos x) - (1 + 2\sin x\cos x)
=−1+2sin⁡xcos⁡x−1−2sin⁡xcos⁡x=−2= -1 + 2\sin x\cos x - 1 - 2\sin x\cos x = -2

f′(x)=−2(sin⁡x−cos⁡x)2f'(x) = \frac{-2}{(\sin x - \cos x)^2}
18Find the derivative of sec⁡x−1sec⁡x+1\dfrac{\sec x - 1}{\sec x + 1}.Show solution
Given: f(x)=sec⁡x−1sec⁡x+1f(x) = \dfrac{\sec x - 1}{\sec x + 1}

Rewrite: f(x)=1cos⁡x−11cos⁡x+1=1−cos⁡x1+cos⁡xf(x) = \dfrac{\dfrac{1}{\cos x} - 1}{\dfrac{1}{\cos x} + 1} = \dfrac{1 - \cos x}{1 + \cos x}

u=1−cos⁡x, u′=sin⁡xu = 1 - \cos x,\ u' = \sin x; v=1+cos⁡x, v′=−sin⁡xv = 1 + \cos x,\ v' = -\sin x

f′(x)=sin⁡x(1+cos⁡x)−(1−cos⁡x)(−sin⁡x)(1+cos⁡x)2f'(x) = \frac{\sin x(1+\cos x) - (1-\cos x)(-\sin x)}{(1+\cos x)^2}
=sin⁡x+sin⁡xcos⁡x+sin⁡x−sin⁡xcos⁡x(1+cos⁡x)2= \frac{\sin x + \sin x\cos x + \sin x - \sin x\cos x}{(1+\cos x)^2}
=2sin⁡x(1+cos⁡x)2= \frac{2\sin x}{(1+\cos x)^2}
19Find the derivative of sin⁡nx\sin^n x.Show solution
Given: f(x)=sin⁡nxf(x) = \sin^n x

Using chain rule:
f′(x)=nsin⁡n−1x⋅cos⁡xf'(x) = n\sin^{n-1}x \cdot \cos x

Answer: f′(x)=nsin⁡n−1xcos⁡xf'(x) = n\sin^{n-1}x\cos x
20Find the derivative of a+bsin⁡xc+dcos⁡x\dfrac{a + b\sin x}{c + d\cos x}.Show solution
Given: f(x)=a+bsin⁡xc+dcos⁡xf(x) = \dfrac{a + b\sin x}{c + d\cos x}

u=a+bsin⁡x, u′=bcos⁡xu = a + b\sin x,\ u' = b\cos x; v=c+dcos⁡x, v′=−dsin⁡xv = c + d\cos x,\ v' = -d\sin x

f′(x)=bcos⁡x(c+dcos⁡x)−(a+bsin⁡x)(−dsin⁡x)(c+dcos⁡x)2f'(x) = \frac{b\cos x(c + d\cos x) - (a + b\sin x)(-d\sin x)}{(c + d\cos x)^2}
=bccos⁡x+bdcos⁡2x+adsin⁡x+bdsin⁡2x(c+dcos⁡x)2= \frac{bc\cos x + bd\cos^2 x + ad\sin x + bd\sin^2 x}{(c + d\cos x)^2}
=bccos⁡x+bd(cos⁡2x+sin⁡2x)+adsin⁡x(c+dcos⁡x)2= \frac{bc\cos x + bd(\cos^2 x + \sin^2 x) + ad\sin x}{(c + d\cos x)^2}
=bccos⁡x+bd+adsin⁡x(c+dcos⁡x)2= \frac{bc\cos x + bd + ad\sin x}{(c + d\cos x)^2}
21Find the derivative of sin⁡(x+a)cos⁡x\dfrac{\sin(x + a)}{\cos x}.Show solution
Given: f(x)=sin⁡(x+a)cos⁡xf(x) = \dfrac{\sin(x+a)}{\cos x}

u=sin⁡(x+a), u′=cos⁡(x+a)u = \sin(x+a),\ u' = \cos(x+a); v=cos⁡x, v′=−sin⁡xv = \cos x,\ v' = -\sin x

f′(x)=cos⁡(x+a)cos⁡x−sin⁡(x+a)(−sin⁡x)cos⁡2xf'(x) = \frac{\cos(x+a)\cos x - \sin(x+a)(-\sin x)}{\cos^2 x}
=cos⁡(x+a)cos⁡x+sin⁡(x+a)sin⁡xcos⁡2x= \frac{\cos(x+a)\cos x + \sin(x+a)\sin x}{\cos^2 x}
=cos⁡(x+a−x)cos⁡2x=cos⁡acos⁡2x=cos⁡asec⁡2x= \frac{\cos(x+a-x)}{\cos^2 x} = \frac{\cos a}{\cos^2 x} = \cos a \sec^2 x
22Find the derivative of x4(5sin⁡x−3cos⁡x)x^4(5\sin x - 3\cos x).Show solution
Given: f(x)=x4(5sin⁡x−3cos⁡x)f(x) = x^4(5\sin x - 3\cos x)

Using product rule: u=x4, u′=4x3u = x^4,\ u' = 4x^3; v=5sin⁡x−3cos⁡x, v′=5cos⁡x+3sin⁡xv = 5\sin x - 3\cos x,\ v' = 5\cos x + 3\sin x

f′(x)=4x3(5sin⁡x−3cos⁡x)+x4(5cos⁡x+3sin⁡x)f'(x) = 4x^3(5\sin x - 3\cos x) + x^4(5\cos x + 3\sin x)
=x3[4(5sin⁡x−3cos⁡x)+x(5cos⁡x+3sin⁡x)]= x^3[4(5\sin x - 3\cos x) + x(5\cos x + 3\sin x)]
=x3[20sin⁡x−12cos⁡x+5xcos⁡x+3xsin⁡x]= x^3[20\sin x - 12\cos x + 5x\cos x + 3x\sin x]
23Find the derivative of (x2+1)cos⁡x(x^2 + 1)\cos x.Show solution
Given: f(x)=(x2+1)cos⁡xf(x) = (x^2 + 1)\cos x

Using product rule: u=x2+1, u′=2xu = x^2+1,\ u' = 2x; v=cos⁡x, v′=−sin⁡xv = \cos x,\ v' = -\sin x

f′(x)=2xcos⁡x+(x2+1)(−sin⁡x)=2xcos⁡x−(x2+1)sin⁡xf'(x) = 2x\cos x + (x^2+1)(-\sin x) = 2x\cos x - (x^2+1)\sin x
24Find the derivative of (ax2+sin⁡x)(p+qcos⁡x)(ax^2 + \sin x)(p + q\cos x).Show solution
Given: f(x)=(ax2+sin⁡x)(p+qcos⁡x)f(x) = (ax^2 + \sin x)(p + q\cos x)

u=ax2+sin⁡x, u′=2ax+cos⁡xu = ax^2 + \sin x,\ u' = 2ax + \cos x; v=p+qcos⁡x, v′=−qsin⁡xv = p + q\cos x,\ v' = -q\sin x

f′(x)=(2ax+cos⁡x)(p+qcos⁡x)+(ax2+sin⁡x)(−qsin⁡x)f'(x) = (2ax + \cos x)(p + q\cos x) + (ax^2 + \sin x)(-q\sin x)
=(2ax+cos⁡x)(p+qcos⁡x)−qsin⁡x(ax2+sin⁡x)= (2ax + \cos x)(p + q\cos x) - q\sin x(ax^2 + \sin x)
25Find the derivative of (x+cos⁡x)(x−tan⁡x)(x + \cos x)(x - \tan x).Show solution
Given: f(x)=(x+cos⁡x)(x−tan⁡x)f(x) = (x + \cos x)(x - \tan x)

u=x+cos⁡x, u′=1−sin⁡xu = x + \cos x,\ u' = 1 - \sin x; v=x−tan⁡x, v′=1−sec⁡2x=−tan⁡2xv = x - \tan x,\ v' = 1 - \sec^2 x = -\tan^2 x

f′(x)=(1−sin⁡x)(x−tan⁡x)+(x+cos⁡x)(1−sec⁡2x)f'(x) = (1 - \sin x)(x - \tan x) + (x + \cos x)(1 - \sec^2 x)
=(1−sin⁡x)(x−tan⁡x)−(x+cos⁡x)tan⁡2x= (1 - \sin x)(x - \tan x) - (x + \cos x)\tan^2 x
26Find the derivative of 4x+5sin⁡x3x+7cos⁡x\dfrac{4x + 5\sin x}{3x + 7\cos x}.Show solution
Given: f(x)=4x+5sin⁡x3x+7cos⁡xf(x) = \dfrac{4x + 5\sin x}{3x + 7\cos x}

u=4x+5sin⁡x, u′=4+5cos⁡xu = 4x + 5\sin x,\ u' = 4 + 5\cos x; v=3x+7cos⁡x, v′=3−7sin⁡xv = 3x + 7\cos x,\ v' = 3 - 7\sin x

f′(x)=(4+5cos⁡x)(3x+7cos⁡x)−(4x+5sin⁡x)(3−7sin⁡x)(3x+7cos⁡x)2f'(x) = \frac{(4 + 5\cos x)(3x + 7\cos x) - (4x + 5\sin x)(3 - 7\sin x)}{(3x + 7\cos x)^2}

Expanding numerator:
=12x+28cos⁡x+15xcos⁡x+35cos⁡2x−(12x−28xsin⁡x+15sin⁡x−35sin⁡2x)= 12x + 28\cos x + 15x\cos x + 35\cos^2 x - (12x - 28x\sin x + 15\sin x - 35\sin^2 x)
=12x+28cos⁡x+15xcos⁡x+35cos⁡2x−12x+28xsin⁡x−15sin⁡x+35sin⁡2x= 12x + 28\cos x + 15x\cos x + 35\cos^2 x - 12x + 28x\sin x - 15\sin x + 35\sin^2 x
=28cos⁡x+15xcos⁡x+35(cos⁡2x+sin⁡2x)+28xsin⁡x−15sin⁡x= 28\cos x + 15x\cos x + 35(\cos^2 x + \sin^2 x) + 28x\sin x - 15\sin x
=28cos⁡x+15xcos⁡x+35+28xsin⁡x−15sin⁡x= 28\cos x + 15x\cos x + 35 + 28x\sin x - 15\sin x

f′(x)=35+28cos⁡x−15sin⁡x+x(15cos⁡x+28sin⁡x)(3x+7cos⁡x)2f'(x) = \frac{35 + 28\cos x - 15\sin x + x(15\cos x + 28\sin x)}{(3x + 7\cos x)^2}
27Find the derivative of x2cos⁡(π4)sin⁡x\dfrac{x^2\cos\left(\dfrac{\pi}{4}\right)}{\sin x}.Show solution
Given: f(x)=x2cos⁡(π4)sin⁡xf(x) = \dfrac{x^2\cos\left(\dfrac{\pi}{4}\right)}{\sin x}

Note: cos⁡(π4)=12\cos\left(\dfrac{\pi}{4}\right) = \dfrac{1}{\sqrt{2}} is a constant. So f(x)=12⋅x2sin⁡xf(x) = \dfrac{1}{\sqrt{2}} \cdot \dfrac{x^2}{\sin x}.

u=x2, u′=2xu = x^2,\ u' = 2x; v=sin⁡x, v′=cos⁡xv = \sin x,\ v' = \cos x

f′(x)=12⋅2xsin⁡x−x2cos⁡xsin⁡2x=x2cos⁡(π4)(2sin⁡x−xcos⁡x)xsin⁡2xf'(x) = \frac{1}{\sqrt{2}} \cdot \frac{2x\sin x - x^2\cos x}{\sin^2 x} = \frac{x^2\cos\left(\dfrac{\pi}{4}\right)(2\sin x - x\cos x)}{x\sin^2 x}

f′(x)=cos⁡(π4)(2xsin⁡x−x2cos⁡x)sin⁡2x\boxed{f'(x) = \frac{\cos\left(\dfrac{\pi}{4}\right)(2x\sin x - x^2\cos x)}{\sin^2 x}}
28Find the derivative of x1+tan⁡x\dfrac{x}{1 + \tan x}.Show solution
Given: f(x)=x1+tan⁡xf(x) = \dfrac{x}{1 + \tan x}

u=x, u′=1u = x,\ u' = 1; v=1+tan⁡x, v′=sec⁡2xv = 1 + \tan x,\ v' = \sec^2 x

f′(x)=1⋅(1+tan⁡x)−x⋅sec⁡2x(1+tan⁡x)2=1+tan⁡x−xsec⁡2x(1+tan⁡x)2f'(x) = \frac{1 \cdot (1 + \tan x) - x \cdot \sec^2 x}{(1 + \tan x)^2} = \frac{1 + \tan x - x\sec^2 x}{(1 + \tan x)^2}
29Find the derivative of (x+sec⁡x)(x−tan⁡x)(x + \sec x)(x - \tan x).Show solution
Given: f(x)=(x+sec⁡x)(x−tan⁡x)f(x) = (x + \sec x)(x - \tan x)

u=x+sec⁡x, u′=1+sec⁡xtan⁡xu = x + \sec x,\ u' = 1 + \sec x\tan x; v=x−tan⁡x, v′=1−sec⁡2x=−tan⁡2xv = x - \tan x,\ v' = 1 - \sec^2 x = -\tan^2 x

f′(x)=(1+sec⁡xtan⁡x)(x−tan⁡x)+(x+sec⁡x)(1−sec⁡2x)f'(x) = (1 + \sec x\tan x)(x - \tan x) + (x + \sec x)(1 - \sec^2 x)
=(1+sec⁡xtan⁡x)(x−tan⁡x)−(x+sec⁡x)tan⁡2x= (1 + \sec x\tan x)(x - \tan x) - (x + \sec x)\tan^2 x
30Find the derivative of xsin⁡nx\dfrac{x}{\sin^n x}.Show solution
Given: f(x)=xsin⁡nxf(x) = \dfrac{x}{\sin^n x}

u=x, u′=1u = x,\ u' = 1; v=sin⁡nx, v′=nsin⁡n−1xcos⁡xv = \sin^n x,\ v' = n\sin^{n-1}x\cos x

f′(x)=1⋅sin⁡nx−x⋅nsin⁡n−1xcos⁡xsin⁡2nxf'(x) = \frac{1 \cdot \sin^n x - x \cdot n\sin^{n-1}x\cos x}{\sin^{2n} x}
=sin⁡n−1x(sin⁡x−nxcos⁡x)sin⁡2nx= \frac{\sin^{n-1}x(\sin x - nx\cos x)}{\sin^{2n} x}
=sin⁡x−nxcos⁡xsin⁡n+1x= \frac{\sin x - nx\cos x}{\sin^{n+1} x}

Stuck on a step?

Ask Super Tutor AI to explain any solution on this page in a simpler way — free, 24x7.

Ask a Doubt Free

Frequently Asked Questions

What are the important topics in Limits and Derivatives for Bihar Board Class 11 Mathematics?
Limits and Derivatives covers several key topics that are frequently asked in Bihar Board Class 11 board exams. Focus on the core concepts listed on this page and practise related questions to build confidence.
How to score full marks in Limits and Derivatives — Bihar Board Class 11 Mathematics?
Understand the core concepts first, then work through the 84 practice questions available for this chapter. Revise formulas and definitions regularly, and use flashcards for quick recall before the exam.
Where can I get free NCERT Solutions for Limits and Derivatives Class 11 Mathematics?
This page has free step-by-step NCERT Solutions for every exercise question in Limits and Derivatives (Bihar Board Class 11 Mathematics) — written the way examiners award marks: given, formula, working, answer.

Sources & Official References

Content is aligned to the official syllabus. Refer to the board website for the latest curriculum.

More resources for Limits and Derivatives

All 14 chapters — Bihar Board Class 11 Mathematics NCERT Solutions

For serious students

Get the full Limits and Derivatives chapter — for free.

Quizzes, flashcards, AI doubt-solver and a step-by-step study plan for Bihar Board Class 11 Mathematics.

Try Super Tutor Free
Try the full chapter — Free